| I never really got into geometry very much, so I often convert geometry
problems to algebra. Place the point mentioned in the problem at (0,0) in the
x-y plane. Put the northwest corner at (3,0). The southwest corner is now
somewhere "above and to the left" of (3,0), and the northeast corner is
somewhere "below and to the left". Let the southwest corner be at (a,b) and
the northeast corner be at (c,d). Because these are corners on a square and
because of their arrangement, if (a,b)-(3,0) = (x,y), then
(3,0)-(c,d) = (y,-x). This gives:
a-3 = -(0-d) and b-0 = 3-c, or
a = d+3 and b = 3-c. (Equations 2 and 3)
The line from the origin to the southwest corner is 5, so
a*a + b*b = 5*5. (Equation 0)
Similarly,
c*c + d*d = 4*4. (Equation 1)
Substituting for a and b in Equation 0 yields:
(d+3)**2 + (3-c)**2 = 25, so
d*d+6d+9 + 9-6c+c*c = 25.
6d+9+9-6c = 25-16 (by subtracting Equation 1),
6d-6c = -9,
d-c = -3/2, and
c = d+3/2. (Equation 4)
Substituting this into Equation 1 gives:
(d+3/2)**2 + d*d = 16,
2*d*d+3d+9/4 = 16, and
2*d*d+3d-55/4 = 0.
The quadratic formula gives d as:
d = [-3 +/- sqrt(9--2*55)] / 4,
= [-3 +/- sqrt(119)] / 4.
The square root of 119 is larger than 3, and we know d must be negative, since
the northeast corner has been placed "below and to the left" of (3,0). So
d = [-3 - sqrt(119)] / 4.
Substituting this into Equation 4 gives:
c = [-3 - sqrt(119)]/4 + 3/2,
= [3 - sqrt(119)] / 4.
Substituting for c and d in Equations 2 and 3 yields:
a = [-3 - sqrt(119)]/4 + 3 and b = 3 - [3 - sqrt(119)] / 4,
a = [9 - sqrt(119)] / 4 and b = [9 + sqrt(119)] / 4.
To find the area of the square, consider the vector from the northwest corner
to the southwest corner, (a,b)-(3,0). Of course, a-3 = d, so the length of
a side is just sqrt(b*b+d*d). So:
area = b*b + d*d,
= [81 + 18 sqrt(119) + 119]/16 + [9 + 6 sqrt(119) + 119]/16,
= [328 + 24 sqrt(119)] / 16,
= [41 + 3 sqrt(119)] / 2.
This is not elegant, but it is not obscure, either. Some insight might be
obtained if one were to graph Equations 0, 1, and 4, along with the equation
one would get by substituting for c and d in Equation 1 instead of a and b
in Equation 0.
-- edp
|
| On the other hand, I was always more comfortable with geometry, so I did this
problem another way.
My geometry teacher was always stressing the importance of drawing
perpendiculars, so I drew lines north and west from that point in the square,
giving me 4 right triangles to work from.
Give the name x to the length of the vertical line, and y to the horizontal.
Call the length of a side, s.
Use the pythagorean theorem on the 4 right triangles to get three equations
(two of the triangles are the same):
x**2 + y**2 = 3**2 (equation 1)
x**2 + (s-y)**2 = 5**2 (equation 2)
(s-x)**2 + y**2 = 4**2 (equation 3)
Visual aid:
x (s-x)
----------------------
|\ | / |
| \ 3 | y / |
y | \ | / 4 | N
| \ | / | W E
|________\ / | S
| x / |
| / |
(s-y)| / |
| /5 |
| / |
|----------------------
(s-y)**2 - y**2 = 16 (equ. 2 - equ. 1)
s**2 - 2*s*y = 16
y = (s**2 - 16)/(2*s)
By starting from equ. 3 - equ. 1, obtain by the same process
x = (s**2 - 7)/(2*s)
Plug these new expressions for x and y back into equ. 1, multiply it out, and
collect the terms. Let a stand for s**2.
(a-7)**2 + (a-16)**2 = 9 * 4a
2*a**2 - 82*a + 305 = 0
The quadratic equation gives
a = (41 +/- 3*sqrt(119))/2
so s = 6.0715... or s = 2.034...
The first one is correct for the problem as stated. The value of s = 2.034..
gives a meeting point of the three lines that is outside the square, which
would be an interesting trick answer, depending on how the question was
phrased.
The .1 and .2 solutions took roughly the same amount of figuring, just choosing
different coordinate axes to work with. Is there some strange and elegant
solution you were thinking of?
-Jim Grant, Littleton MA.
|
| Well \my/ geometry teacher always stressed the importance of
group theory and transformations (well, not really - but it sounds
good), so my favorite solution to this problem goes as follows:
Let the square be ABCD (counterclockwise) with P inside and
PA=4, PB=3, PC=5. (BA is horizontal at the top.)
Since for any rectangle and any point P, we have
PA^2+PC^2=PB^2+PD^2 (which can be seen by dropping those perpendiculars),
therefore PD=sqrt(32).
Now rotate triangle ABP 90 degrees clockwise around point B so that A goes
into C and P goes into a point A' (to the left of BC). Then rotate triangle
PCD 90 degrees clockwise around point D so that C goes into A and P goes
into a point B' (to the right of AD).
BA'C is congruent to BAP, so A'C=4. PBA' is an isosceles right triangle
so PA'=3sqrt(2). Similarly, AB'D is congruent to CPD so AB'=5
and triangle PDB' is an isosceles right triangle, so PB'=sqrt(32)sqrt(2)=8.
Let s be the side of the square, so that its area is s^2.
Let [XYZ] denote the area of triangle XYZ.
Let [a,b,c] denote the area of a triangle with sides of length a, b, c.
The area of the square is also equal to [PAB]+[PBC]+[PCD]+[PDA]=
[PBC]+[BCA']+[PAD]+[ADB']=[PBA']+[PCA']+[APB']+[DPB']=
3^2/2+[4,5,3sqrt(2)]+[sqrt(32)^2/2+[4,5,8].
These areas are easily found by Heron's formula, so that gives us
the value of s^2 from which we can get s.
|
| There was a recent article on a generalization of this problem
to regular n-gons, but I have been searching for it for a week now
and haven't been able to come up with it, so I'll try to paraphrase
the result from memory:
Let A, B, C be consecutive vertices of a regular n-gon.
Let P be inside such that PA=4, PB=3, PC=5. Find the side of the n-gon.
Solution: Let theta = angle ABC be the (known) angle of a regular n-gon.
Rotate triangle ABP around point B through an angle theta
so that A coincides with C and P transforms into a point Q
outside of the n-gon. Triangle ABP is congruent to triangle CBQ
so CQ=4. Draw in PQ. Angle PBQ is known (theta) so one can find the
length of PQ by the law of cosines in triangle PQ. Another law of cosines
gives you the cosines of angles QPB and QPC. Thus you can also get
the sines of these angles. Using the formula for the cosine of the
sum of two angles then gives you the cosine of angle BPC, from which
a final application of the law of cosines (in triangle PBC) gives you
BC, the desired length of the side of the n-gon.
|
| My solution is related to Stan's but comes in fewer pieces, and yields results
that relate well to the algebraic solutions in .1 and .2:
Using Stan's notation, rotate APB about A -90 degrees, and rotate BPC about
C +90 degrees. The rotated segments passing trough D are colinear and form
a sigle segment of length 6. The pentagon thus formed can be dissected into
two isoceles right triangles of sides 4 and 5, resp., and a triangle of sides
4*sqrt(2), 5*sqrt(2), and 6.. The areas are easy to calculate from Heron's
formula.
The rational part of the solution, namely 41/2, is exactly the area of the
two isoceles right triangles. Therefore the irrational part is exactly the
area of the third triangle in the dissection.
Another method of solution is to equate the areas (by Heron's formula, again)
of APB+BPC+APC (i.e. half the square) with ABC. This results in a lot of
hairy algebra but something like VAXSYMA will resolve it quickly.
Lynn Yarbrough
|
| Back in 1984 I attended a Symbolic Math Conference at NYU, where several
systems such as MACSYMA, MAPLE, REDUCE, etc. were on display. I showed this
problem to several people and got a few solutions. The MAPLE people
remembered it and put it in the [.DEMO.RESEARCH] subdirectory of the MAPLE
4.3 distribution, as example '_NYU'. The problem statement and ALL the
solutions follow:
read verifier:
#
# From a user at the N.Y.U conference 1984.
#
# verifier: 1 solution with [0] free variables in 12.02 secs (1/Jun/87)
#
eqn1 := x^2 + y^2 = 9;
eqn3 := x^2 + (s-y)^2 = 25;
eqn2 := (s-x)^2 + y^2 = 16;
eqns := {eqn.(1..3)};
vars := {x,y,s};
printlevel := 4;
st := time(); soln := [solve(eqns,vars)]; tt := time()-st;
verifier(soln,eqns);
done
2 2
- 7 + %1 - 16 + %1
soln := [{s = %1, x = 1/2 ---------, y = 1/2 ----------}]
%1 %1
%1 :=
2 4
RootOf(305 - 82 _Z + 2 _Z )
|